제출 #506084

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
506084		aryan12	A Difficult(y) Choice (BOI21_books)	C++17	100 / 100	2 ms	1056 KiB

books

#include <bits/stdc++.h>
#include "books.h"
using namespace std;
//
// --- Sample implementation for the task books ---
//
// To compile this program with the sample grader, place:
//     books.h books_sample.cpp sample_grader.cpp
// in a single folder and run:
//     g++ books_sample.cpp sample_grader.cpp
// in this folder.
//

void solve(int N, int K, long long A, int S) {
    long long L = 1, R = N;
    long long omk = 0;
    long long ans[N + 1];
    for(long long i = 0; i < N + 1; i++) {
        ans[i] = -1;
    }
    while(L <= R) {
        long long mid = (L + R) >> 1;
        ans[mid] = skim(mid);
        if(ans[mid] <= A) {
            L = mid + 1;
            omk = mid;
        }
        else {
            R = mid - 1;
        }
    }
    for(long long i = 1; i <= K; i++) {
        //cout << "i = " << i << "\n";
        if(ans[i] == -1) {
            ans[i] = skim(i);
        }
    }
    //cout << "omk = " << omk << ", K = " << K << "\n";
    for(long long i = omk; i > max(omk - K, 0LL); i--) {
        //cout << "i = " << i << "\n";
        if(ans[i] == -1) {
            ans[i] = skim(i);
        }
    }
    long long value = 0;
    vector<int> res;
    set<long long> res2;
    for(long long i = 1; i <= K; i++) {
        value += ans[i];
        res.push_back(i);
        res2.insert(i);
    }
    if(value >= A && value <= 2LL * A) {
        answer(res);
        return;
    }
    long long cnt = 1;
    for(long long i = omk; i > max(omk - K, 0LL); i--) {
        if(res2.count(i)) 
            continue;
        res2.erase(cnt);
        res2.insert(i);
        res.erase(res.begin());
        res.push_back(i);
        value -= ans[cnt];
        value += ans[i];
        cnt++;
        if(value >= A && value <= 2LL * A) {
            answer(res);
            return;
        }
    }
    if(omk == N || omk < K + 1) {
        impossible();
        return;
    }
    ans[omk + 1] = skim(omk + 1);
    res.clear();
    value = 0;
    for(long long i = 1; i < K; i++) {
        res.push_back(i);
        value += ans[i];
    }
    res.push_back(omk + 1);
    value += ans[omk + 1];
    if(value >= A && value <= 2LL * A) {
        answer(res);
        return;
    }
    impossible();
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 10 / 10
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 20 / 20
• Subtask 7: 20 / 20